Poj 2478-Farey Sequence 欧拉函数,素数,线性筛

Farey Sequence

Time Limit: 1000MS		Memory Limit: 65536K
Total Submissions: 14291		Accepted: 5647

Description

The Farey Sequence Fn for any integer n with n >= 2 is the set of irreducible rational numbers a/b with 0 < a < b <= n and gcd(a,b) = 1 arranged in increasing order. The first few are 
F2 = {1/2} 
F3 = {1/3, 1/2, 2/3} 
F4 = {1/4, 1/3, 1/2, 2/3, 3/4} 
F5 = {1/5, 1/4, 1/3, 2/5, 1/2, 3/5, 2/3, 3/4, 4/5}

You task is to calculate the number of terms in the Farey sequence Fn.

Input

There are several test cases. Each test case has only one line, which contains a positive integer n (2 <= n <= 10⁶). There are no blank lines between cases. A line with a single 0 terminates the input.

Output

For each test case, you should output one line, which contains N(n) ---- the number of terms in the Farey sequence Fn. 

Sample Input

2
3
4
5
0

Sample Output

1
3
5
9

Source

POJ Contest,Author:Mathematica@ZSU

题解：

直接欧拉函数，求个前缀和，就好了。

 #include<cstdio>
 #include<cstring>
 #include<cstdlib>
 #include<cmath>
 #include<iostream>
 #include<algorithm>
 using namespace std;
 #define LL long long
 LL qz[];
 int phi[],prime[],tot;
 bool vis[];
 void Eular()
 {
     int i,j;
     phi[]=;tot=;
     for(i=;i<=;i++)
     {
         if(vis[i]==false)
         {
             prime[++tot]=i;
             phi[i]=i-;
         }
         for(j=;j<=tot&&prime[j]*i<=;j++)
         {
             vis[prime[j]*i]=true;
             if(i%prime[j]==)
             {
                 phi[prime[j]*i]=phi[i]*prime[j];
                 break;
             }
             phi[prime[j]*i]=phi[prime[j]]*phi[i];
         }
     }
 }
 void Qz()
 {
     for(int i=;i<=;i++)qz[i]=qz[i-]+phi[i];
 }
 int main()
 {
     int n;
     Eular();
     //for(int i=1;i<=100;i++)printf("%d ",phi[i]);
     //printf("\n");
     Qz();
     while()
     {
         scanf("%d",&n);
         if(n==)break;
         printf("%lld\n",qz[n]-);
     }
     return ;
 }